jmc

To get a handle on the constant $t,$ we can look at some particular cases.

Suppose we let $AB$ approach a vertical line. Then $\frac{1}{AC}$ approaches 0, and $B$ approaches $(0,0),$ so $\frac{1}{AC}+\frac{1}{BC}$ approaches $c.$ Hence, $$t=\frac{1}{c}.$$Now, suppose we take $A=(\sqrt{c},c)$ and $B=(-\sqrt{c},c).$ Then $$t=\frac{1}{AC}+\frac{1}{BC}=\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{c}}=\frac{2}{\sqrt{c}}.$$Hence, $\frac{1}{c}=\frac{2}{\sqrt{c}},$ so $\sqrt{c}=\frac{1}{2},$ and $c=\frac{1}{4}.$ Therefore, $t=\boxed{4}.$ (Note that this makes $C$ the focus of the parabola.)

For a full solution, let's check that this value works. Let $y=mx+\frac{1}{4}$ be the equation of line $AB.$ Setting $y=x^2,$ we get $$x^2=mx+\frac{1}{4},$$or $x^2-mx-c=0.$ Let $x_1$ and $x_2$ be the roots of this equation. By Vieta's formulas, $x_1+x_2=m$ and $x_1{x}_2=-\frac{1}{4}.$

Also, $A$ and $B$ are $(x_1,x_1^2)$ and $(x_2,x_2^2)$ in some order, so $$\begin{array}{rl}\frac{1}{AC}+\frac{1}{BC}& =\frac{1}{\sqrt{x_1^2+(x_1^2-\frac{1}{4}{)}^2}}+\frac{1}{\sqrt{x_2^2+(x_2^2-\frac{1}{4}{)}^2}}\\ & =\frac{1}{\sqrt{x_1^2+x_1^4-\frac{1}{2}{x}_1^2+\frac{1}{16}}}+\frac{1}{\sqrt{x_2^2+x_2^4-\frac{1}{2}{x}_2^2+\frac{1}{16}}}\\ & =\frac{1}{\sqrt{x_1^4+\frac{1}{2}{x}_1^2+\frac{1}{16}}}+\frac{1}{\sqrt{x_2^4+\frac{1}{2}{x}_2^2+\frac{1}{16}}}\\ & =\frac{1}{\sqrt{(x_1^2+\frac{1}{4}{)}^2}}+\frac{1}{\sqrt{(x_2^2+\frac{1}{4}{)}^2}}\\ & =\frac{1}{x_1^2+\frac{1}{4}}+\frac{1}{x_2^2+\frac{1}{4}}.\end{array}$$We have that $x_1^2{x}_2^2=(x_1{x}_2{)}^2={\left(-\frac{1}{4}\right)}^2=\frac{1}{16}$ and $$x_1^2+x_2^2=(x_1+x_2{)}^2-2x_1{x}_2=m^2+\frac{1}{2}.$$Hence, $$\begin{array}{rl}\frac{1}{x_1^2+\frac{1}{4}}+\frac{1}{x_2^2+\frac{1}{4}}& =\frac{x_1^2+\frac{1}{4}+x_2^2+\frac{1}{4}}{(x_1^2+\frac{1}{4})(x_2^2+\frac{1}{4})}\\ & =\frac{x_1^2+x_2^2+\frac{1}{2}}{x_1^2{x}_2^2+\frac{1}{4}(x_1^2+x_2^2)+\frac{1}{16}}\\ & =\frac{m^2+1}{\frac{1}{16}+\frac{1}{4}(m^2+\frac{1}{2})+\frac{1}{16}}\\ & =\frac{m^2+1}{\frac{1}{4}{m}^2+\frac{1}{4}}\\ & =4.\end{array}$$